Problem: Multiply the following complex numbers: $({2i}) \cdot ({1+3i})$
Answer: Complex numbers are multiplied like any two binomials. First use the distributive property: $ ({2i}) \cdot ({1+3i}) = $ $ ({0} \cdot {1}) + ({0} \cdot {3}i) + ({2}i \cdot {1}) + ({2}i \cdot {3}i) $ Then simplify the terms: $ (0) + (0i) + (2i) + (6 \cdot i^2) $ Imaginary unit multiples can be grouped together. $ 0 + (0 + 2)i + 6i^2 $ After we plug in $i^2 = -1$ , the result becomes $ 0 + (0 + 2)i - 6 $ The result is simplified: $ (0 - 6) + (2i) = -6+2i $